The displacement x in meter of a particle of mass m kg moving in one direction under the action of a force is related to the time t in second by the equation : 

\(x = (t - 3)^2\)

The work done by the force (in joules) in first six seconds is : 

0

According to ques : \(x = (t-3)^2\)

\(x = t^2 - 6t + 9\)

\(v = \frac{dx}{dt} = 2t - 6\)

At t = 0, v = 2 x 0 - 6 = -6

at t = 6 s, v = 2 x 6 - 6 = 6

Initial and final KE are same,

From Work Energy Theorem , Work Done is equal to change in K.E.

$W = \Delta K$

hence no work is done.